Starting Out with C++: Early Objects (9th Edition)
9th Edition
ISBN: 9780134400242
Author: Tony Gaddis, Judy Walters, Godfrey Muganda
Publisher: PEARSON
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Expert Solution & Answer
Chapter 19, Problem 2RQE
Program Description Answer
A binary tree contains left and right pointers points to the “root” node.
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#ifndef BT_NODE_H#define BT_NODE_H
struct btNode{ int data; btNode* left; btNode* right;};
// pre: bst_root is root pointer of a binary search tree (may be 0 for// empty tree) and portArray has the base address of an array large// enough to hold all the data items in the binary search tree// post: The binary search tree has been traversed in-order and the data// values are written (as they are encountered) to portArray in// increasing positional order starting from the first elementvoid portToArrayInOrder(btNode* bst_root, int* portArray);void portToArrayInOrderAux(btNode* bst_root, int* portArray, int& portIndex);
// pre: (none)// post: dynamic memory of all the nodes of the tree rooted at root has been// freed up (returned back to heap/freestore) and the tree is now empty// (root pointer contains the null address)void tree_clear(btNode*& root);
// pre: (none)// post: # of nodes contained in tree rooted at root is returnedint…
Data Structures
501324-3
Sections: 2759& 3137
Question : Tree
Draw the Binary Search Tree that would result from the insertion of the following integer keys:
10
6
12
8
16
24
2
5
b) After you create the binary search tree, Show the output of:
1) Inorder Tree Traversal,
2) Preorder Tree Traversal,
3) Postorder Tree Traversal,
Find the errors in the program then correct them.
CODE:
#include <bits/stdc++.h>using namespace std; /* A binary tree node has key, pointer to leftchild and a pointer to right child */struct Node { int key; struct Node *left, *right}; /* function to create a new node of tree andreturn pointer */struct Node* newNode(int key){ struct Node* temp = new node; temp->key = key; temp->left = temp->right = NULL; return temp;}; /* Inorder traversal of a binary tree*/void inorder(struct Node* temp){ if (!temp) return 0 inorder(temp->left); cout << temp->key << " "; inorder(temp->right)} /* function to delete the given deepest node(d_node) in binary tree */void deletDeepest(struct Node* root, struct Node* d_node){ queue<struct Node*> q q.push(root); // Do level order traversal until last node struct Node* temp; while (!q.empty()) { temp = q.front(); q.pop(); if (temp…
Chapter 19 Solutions
Starting Out with C++: Early Objects (9th Edition)
Ch. 19.1 - Prob. 19.1CPCh. 19.1 - Prob. 19.2CPCh. 19.1 - Prob. 19.3CPCh. 19.1 - Prob. 19.4CPCh. 19.1 - Prob. 19.5CPCh. 19.1 - Prob. 19.6CPCh. 19.2 - Prob. 19.7CPCh. 19.2 - Prob. 19.8CPCh. 19.2 - Prob. 19.9CPCh. 19.2 - Prob. 19.10CP
Ch. 19.2 - Prob. 19.11CPCh. 19.2 - Prob. 19.12CPCh. 19 - Prob. 1RQECh. 19 - Prob. 2RQECh. 19 - Prob. 3RQECh. 19 - Prob. 4RQECh. 19 - Prob. 5RQECh. 19 - Prob. 6RQECh. 19 - Prob. 7RQECh. 19 - Prob. 8RQECh. 19 - Prob. 9RQECh. 19 - Prob. 10RQECh. 19 - Prob. 11RQECh. 19 - Prob. 12RQECh. 19 - Prob. 13RQECh. 19 - Prob. 14RQECh. 19 - Prob. 15RQECh. 19 - Prob. 16RQECh. 19 - Prob. 17RQECh. 19 - Prob. 18RQECh. 19 - Prob. 19RQECh. 19 - Prob. 20RQECh. 19 - Prob. 1PCCh. 19 - Prob. 2PCCh. 19 - Prob. 3PCCh. 19 - Prob. 4PCCh. 19 - Prob. 5PCCh. 19 - Prob. 6PCCh. 19 - Prob. 7PCCh. 19 - Prob. 8PCCh. 19 - Prob. 9PCCh. 19 - Prob. 10PC
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